CHEMMMM
Matter
-> Pure Substance
-> Mixture -> Heterogeneous mixture and Homogeneous mixture
A solution is defined as a homogeneous mixture of two or more substances in a single phase.
Homogenous means that the mixture appears uniform because the particles of both substances are evenly interspersed.
The phase of a solution may be solid, liquid, or gas.
Components of a Solution
Solute: The substance being dissolved.
Solvent: The dissolving medium, usually present in greater amounts.
The process of dissolving is called solvation, or when water is the solvent, hydration.
-> STEP 1: Solvent is attracted to the solute.
-> STEP 2: Solvent particles surround the solute particles and pull them into the solution.
Solubility is defined as the maximum amount of solute that can be dissolved in a specific amount of solvent at a specific temperature.
An unsaturated solution contains less than the maximum amount of solute. Additional solutes will dissolve.
A saturated solution contains a maximum amount of solute. Additional solutes won’t dissolve.
A supersaturated solution contains more than the maximum amount of solute and is unstable. Additional solute causes rapid crystallization.
Temperature
Solubility curves show how the solubility of a solute changes with temperature.
At High temperatures:
-> solids are more soluble
-> Gasses are less soluble
Example:
Warm sodas fizz more than cold sodas when you open them because CO2 gas is less soluble at higher temperatures.
Pressure:
Henry’s Law: Gases become more soluble as the pressure above the solution increases.
Aqueous Solution: A solution in which water is solvent.
Electrolytes, Nonelectrolytes
Strong electrolytes separate into ions making solutions that conduct electricity
Weak electrolytes produce a few ions
Nonelectrolytes produce molecules, not ions
Just as different factors affect the amount of solute that dissolves in a solution, there are also various factors that influence how quickly a solute dissolves.
The factors that affect the rate of dissolution are temperature, pressure, agitation, and surface area.
Temperature
Solutes dissolve faster at high temperatures
At high temperatures, solvent particles move faster and solvation occurs more quickly
Everyday Example
When making sweetened iced tea, it is much easier to add the sugar while the tea is still hot. The hot temperature helps the sugar dissolve more quickly.
Agitation
Solutes dissolve faster when the solution is agitated by stirring and shaking.
The solvent immediately surrounding the solute can quickly become saturated. Agitation helps bring fresh solvent into contact with the surface of the solute so that more solute can dissolve.
Everyday Example
When adding sugar to coffee, stirring helps the sugar dissolve faster.
Surface Area
Solutes dissolve faster when the surface area of the solute is increased by crushing it into smaller pieces.
Solvation occurs at the surface of the solute. The greater the surface area, the more opportunities there are for the solvent to attack the solute.
Everyday Example
A sugar cube takes longer to dissolve in a cup of tea than an equal amount of granulated sugar.
Concentration and Dilutions
It's common to change the concentration of a solution.
This is achieved by changing the amount of the solvent
The amount of solute does not change
“Dilution” - The addition of solvent, which decreases the concentration of the solute in the solution
“Concentration” - The removal of solvent, which increases the concentration of the solute in the solution
Calculating Dilutions:
M1V1 = M2V2
M1 = initial molarity
V1 = initial volume
M2 = final molarity
V2 = final volume
In the problems asking for steps to make a dilution:
You need M1 and deltaV (V2-V1)
Steps include:
Take V1 of the M2 solution
Add deltaV of water to the V1 of M1 solution
Percent Composition
Percent Composition is the amount of each element present in the compound as a percentage out of 100%.
Add H total + O total = 18,006 g/mole = molar mass of water
Hydrogen total mass / Compound molar mass = % composition
(2.016 g/mole / 18.006 g/mole) x 100
11.19%
(Mass of element / mass of compound) x 100
Example:
What is the % Composition of Sucrose (C12H22O11) ?
Carbon - 12.011(12) = (144.132 / 342.297) x 100 = 42.107 %
Hydrogen - 1.008(22) = (22.176 / 342.297) x 100 = 6.48 %
Oxygen - 15.999(11) = (175.989 / 342.297) x 100 = 51.41 %
TOTAL - 342.297
Example:
C10H14N2
Carbon - 12.011(10) = (120.11 / 162.236) x 100 = 74.034 %
Hydrogen - 1.008(14) = (14.112 / 162.236) x 100 = 8.698 %
Nitrogen - 14.007(2) = (28.014 / 162.236) x 100 = 17.267 %
TOTAL - 162.236
Empirical Formulas
Tells you the relative ratios of different atoms in a compound in the simplest and most reduced ratio of elements in a compound
Molecular Formulas
Indicates the exact number of atoms in a molecule
Empirical formula from % Composition
Assume 100.0 grams of the compound (changes % to grams)
Convert grams to moles for each element using molar mass
Divide each mole answer by the smallest mole answer
If you do not have all whole numbers at the end, multiply by the smallest whole number to get a whole # ratio
Skip #4 if all numbers are whole numbers. These are your element subscripts.
Example:
What is the empirical formula for a compound that is 21.2% nitrogen, 6.10% hydrogen, 24.3% sulfur, and 48.4% oxygen?
N: 21.2 g / 14.007 = 1.51 / 0.76 = 2
H: 6.10 g / 1.008 = 6.05 / 0.76 = 7.961 or 8
S: 24.3 g / 32.07 = 0.76 / 0.76 = 1
O: 48.4 g / 15.999 = 3.03 / 0.76 = 4
N2H8SO4 - Empirical formula
Example:
What is the empirical formula for a compound that is 69.8% iron and 30.2% oxygen?
Fe: 69.8 g / 55.845 = 1.249 / 1.249 = 1 x 2
O: 30.2 g / 15.999 = 1.887 / 1.249 = 1.512 x 2 = 3
Fe2O3 - Empirical formula
Molecular Formula to Empirical Formula
Determine the Empirical Formula
Use the Formula: Molar mass of Molecular Formula / Molar mass of Empirical Formula
Multiply the subscripts in the empirical formula by this whole number to get the molecular formula
Example:
What is the molecular formula for a compound that is 74.0% carbon, 8.70% hydrogen, and 17.3% nitrogen with a molar mass of 243.39 grams?
C: 74.0 g / 12.011 = 16.161 / 1.23 = 5 (empirical #) x 12.011 (molar mass) = 60.055
H: 8.70 g / 1.008 = 8.631 / 1.23 = 7 (empirical #) x 1.008 (molar mass) = 7.056
N: 17.3 g /14.007 = 1.235 / 1.23 = 1 (empirical #) x 14.007 (molar mass) = 14.01
Empirical formula:
C5H7N1 (weighs 81.07 grams based on the periodic table)
Molecular formula:
243.39 g/mol (molar mass) / 81.07 g/mol (empirical mass) = 3 x (C5H7N)
Therefore, molecular formula = C15H21N3
What is a Chemical Equation?
A chemical equation is written representation of a chemical reaction
List reactants on the left side of the reaction arrow and products on the right side
5 Main types of reactions
Synthesis: Love at first sight
Decomposition: The big break-up
Single replacement: My best friend’s girlfriend
Double replacement: Revenge
Combustion: The player
Synthesis - “Love at first sight”
Basic equation: A+B -> AB
The compound produced is larger and more complex than the reactants
Example: 2H2 + O2 -> 2H2O
Decomposition -> “The Big Break-Up”
Basic Equation: AB -> A+B
The breakdown of a larger compound
Example: 2H2O -> 2H2 + O2
Single Replacement - “My Best Friends Girlfriend”
“Free” element must be more reactive element for a reaction to occur
The further apart 2 elements are on the activity series, the more likely the higher one will replace the lower one in a compound.
Example: 2HCl + K -> H2 + 2KCl
Double Replacement - “Revenge”
Always involves the formation of a molecular compound such as water or either a precipitate or a gas
Example: 2 Kl (aq) + Pb(NO3) 2(aq) -> 2KNO 3(aq) + Pbl 2(s)
Combustion - “The Player”
Produce oxide compounds and energy (very exothermic)
Example: CH4 + 2O2 -> CO2 + 2H2O + ENERGY
Combustion Reactions
Occurs when oxygen reacts with a hydrocarbon to produce water and carbon dioxide.
A hydrocarbon is a compound containing only carbon, hydrogen, and sometimes oxygen.
Example: C10H8(S) + 12O2(g) -> 10CO2(g) + 4H2O(g)
Synthesis (or Combustion)
The combustion of 2 or more substances to form a compound
Decomposition
A compound breaks down into 2 or more simpler substances
Single replacement
Occurs when a single element reacts with an ionic compound and switches place with one of the elements in the compound.
Double replacement
Ions in two compounds “change partners”
Cation of one compound combines with anion of the other compound
Balancing and writing Reactions
The number of atoms on the left of the arrow should be equal to the number of atoms on the right of the arrow. The product mass must be equal to the reactant mass.
Reactants | Products |
1 C | 1 C |
4 H | 2 H |
2 O | 3 O |
CH4 + O2 → CO2 + H2O
Write a t-table for the number of atoms on both sides of the equation. Start by writing down the number of all atoms present.
Coefficients of 1 are “understood” and never written.
Multiply each atom type by the lowest common multiplier that would result in equal amounts on both sides of the reaction. Remember, if atoms are in a compound, the multiplier/coefficient number will affect all atoms of the compound.
Balance metal atoms first.
Balance any groups as a whole group if they remain unchanged on both sides of the equation.
Balance C, H, O last. C before H before O.
Check your work. Make sure you do not have any decimal coefficients. Make sure your coefficients are the most reduced possible.
Phases in reactions:
Phases in a reaction = solid (s), liquid (l), gas (g), and aqueous (aq)
Electrolytes:
Water is the universal solvent and dissolves more substances than any other liquid!
Ionic compounds dissociate or break apart into ions in solution. These are called “electrolytes”. Electrolytes can conduct electrical current due to being charged species.
Covalent compounds even if able to dissolve in water cannot break apart into ions, and do not form electrolytes in water.
Double-displacement Reactions
Double Replacement reactions are not classified as redox reactions as elements do not transfer electrons.
Reactions occur as a more reactive carbon will bond preferentially with a different anion (switch partners)
In the example, Potassium (K) is more reactive and “steals” the NO anion. Lead is left to bond with Iodine
Particle Diagrams for Precipitation
Ionic compounds should be shown as ions in solution or electrolytes
Water should be shown in aqueous solutions
All compounds are labeled (and charge, if needed, is shown!)
An insoluble precipitate should be shown at the bottom of a particle diagram, separated from the water mixture.
Precipitation Reactions
Precipitation Reactions are those that create insoluble solids because of the reaction
Net Ionic Equations
Write the complete ionic equation or reaction
Separate (aq) ionic compounds into respective ions.
Cancel out any ions that were dissolved aqueously on both sides of the reaction. These are spectator ions.
Final net ionic equation shows the formation of the insoluble precipitate.
Reaction: FeCl3 + 3KOH → Fe(OH)3 + 3KCl
The first postulate states that ideal gas particles travel in straight lines until they collide with the walls of their container or each other. These collisions are perfectly elastic.
The second postulate states that the volume of ideal gas particles is negligible compared to the space in between the particles.
The third postulate states that ideal gas particles have no IMFs and do not attract or repel each other.
The fourth postulate states that the average kinetic energy of ideal gas particles is directly proportional to the temperature of the gas.
An ideal gas is gas that follows all postulates of kinetic molecular theory, whereas a real gas is a gas that breaks the postulates of kinetic molecular theory by having, for instance inelastic particle collisions, particles with non-negligible volume, and IMFs between particles.
Boyle’s law states that, for a gas in a sealed, flexible container, as volume increases, pressure will decrease. This is a(n) inverse relationship. This is because gas particles will collide with the walls of the container less often. P1V1 = P2V2
Charle’ law states that, for a gas in a sealed, flexible container, as temperature increases, volume will increase. This is a(n) direct relationship. This is because gas particles will collide with the walls of the container more often. V1/T1 = V2/T2
Gay-Lussac’s law states that, for a gas in a sealed, rigid container, as temperature increases, pressure will increase. This is a(n) direct relationship. This is because gas particles will collide with the walls of the container more often. P1/T1 = P2/T2
Avogadro’s law states that, for a gas in an open, flexible container, as more gas is added to the container, volume will increase. This is a(n) direct relationship. N1/V1 = N2/V2
Stoichiometry: The study of quantities as it relates to chemical reactions.
Balanced chemical equations can be interpreted in many ways: # of Molecules, Mass, AND MOLES (NEW!!)
Mole-Mole Calculations: The coefficients in the balanced equations represent the smallest whole number mole ratio between reactants and products.
Example of mole ratios for the reaction:
1N2(g) + 3H2(g) = 2NH3(g)
1 mol N2/ 2 mol NH3 3 mol H2/1 mol N2 2 mol NH3/ 3 mol H2
Example: If 3.86 moles of potassium (K) reacts completely with excess water (H20), how many moles of hydrogen would be produced?
Potassium + Hydrogen peroxide ( CANCEL WATER) = potassium hydroxide + hydrogen.
2K + 2 H(OH) = 2K(OH) + 1H2
K: 3.86 mol K
U: ? mol H2
3.86 mol K/ 1 * 1 mol H2/ 2 mol K = 1.93 mol H2
Example: How many moles of aluminum will react with 0.512 moles of hydrogen chloride?
Aluminum + Hydrogen chloride = Aluminum chloride + hydrogen
2 Al + 6HCl = 2 AlCl3 + 3 H2
K: 0.512 mol HCl
U: ? mol Al